Best explanation I've seen, the way you explained everything together, I think it finally clicked into the right place in my mind, thanks very much Sam.
Prof. - please clarify... !! 1) At 7:54 you define the internal LG as S'f/S'e. But that completely ISOLATES the summing element (error amplifier), removing it from LG, no? I realize that the lab method - as you show later - breaks the loop externally, and that - magically! - restores the internal error amplifier's presence in the measurement. So, the Q is: why does the "textbook" definition apparently remove the error amp as I mentioned. 2) Is S'in the same as Vref ?? I'd just say - The lower diagram, with B, etc., doesn't match terminology very well with the top one, which uses S', etc. Being consistent with node names, etc., would help!
You missed the sentences following 7:54 : "it is hard to identify...." . The circle at the op amp input is illustrative, it is NOT the summing error junction.
I was wondering if you could remind me of something I can't find the rationale for.... I've read/heard it stated that the Fc crossover (0dB) must be such that Fc
Hi Professor.. can you please make a video on . When to calculate phase margin with respect 0 degrees and 180 degrees... It's very confusing atleast to me...
you have """open loop bandwidth 8.33Khz"" not ""close loop bandwidth"" , is it ?? @ 25:21 close loop bandwidth shows how good this can compensate for load variation , if we have a load that changes around 50Khz , dose it still enable to make regulation at that frequency ?? how we can measure close loop bandwidth for switch mode converter ?
HI Paul. This is the (closed) loop-gain as in the title of slide 49. Then you have the open loop gains, and closed loop gains. For these you need to specify what is input and what is output. E.g. Gain between input to modulator and output. Or: closed loop gain between disturbance at the power input to output. So what do you mean by :" close loop bandwidth for switch mode converter "? Between what and what?
@@sambenyaakov Hi Sam , I see in slide 49 , it is written "loop gain " more exact and accurate is (closed) loop-gain. so this means close loop bandwidth between output to input is 8 Khz when the gain reach 0dB , so my question is if you have a load at output with 50Khz transient switching then this converter can't regulate it at that rate ? is it ? regulation bandwidth is not enough !
Try the RidleyBox instead. Same price, but includes: 4-Channel Frequency Response Analyzer 4-channel Scope Computer Injection Isolator Lifetime RidleyWorks License Watch this video to see how design, implement in hardware AND measure a flyback all in hour. ridleyengineering.com/videos-e/300-design-and-build-a-current-mode-controller-in-one-hour-webinar.html Learn more about the RidleyBox at: www.ridleybox.com
Thank you for the few tips in this video. I you ever have some time to answer my possibly silly questions, I have two : - Is it a problem if the compensation is digital ? For an Infineon PMIC ic, I once asked the field engineer where to get the feedback signal in order to perform loop gain measurement (the datasheet internal schematic was not clear). He answered me that it is not possible to make such measurement because of digital loop. I've never understood this answer and don't see the reason why it would not be possible (an incorrect behavior when faced to a sinusoidal signal, maybe ?). I asked him why, and I'm still waiting for the answer since more than one year ^_^. - Does it makes sense to perform the measurement with the "real load", i.e. on the end application board (different types of load, and continuously moving), rather than with loading resistors ? It seems weird to me to make such measurement (loop gain is load dependant). But a collegue of mine - an old experimented guy - always do this with rather good results (i.e. quickly get an optimized transient response). Thanks again for being so active since many years with so much interesting subjects to share with us.
Thank inico. -No problem to measure the loopgain on a digitally controlled converter. No need to get outside the controller. You can use the method I am showing in this video, injection in the feedback path. Always connected the excitation signal to a low impedance point like the output cap. - Strictly speaking the measurement should be done with the real load. However a resistor is OK in most cases except for the case in which your inverter/converter is feeding another converter which might exhibit a negative dynamic resistance.
Best explanation I've seen, the way you explained everything together, I think it finally clicked into the right place in my mind,
thanks very much Sam.
Thanks. Comments like yours keep me going😊
Your channel and your video are my favorite source of teaching and learning in Power Electronic. Thank you very much Prof. SAM.
Finally a clear & short video about bode100 - thank you Prof!
Thanks
Thank you Professor! Very well explained.
Thanks
great work! thank you very much!
Thanks.
Hi, Is there any alternative methods available to check power supply stability (particularly when we can't afford expensive equipment)
One possibility is with a load step.
Prof. - please clarify... !!
1) At 7:54 you define the internal LG as S'f/S'e. But that completely ISOLATES the summing element (error amplifier), removing it from LG, no?
I realize that the lab method - as you show later - breaks the loop externally, and that - magically! - restores the internal error amplifier's presence in the measurement.
So, the Q is: why does the "textbook" definition apparently remove the error amp as I mentioned.
2) Is S'in the same as Vref ??
I'd just say - The lower diagram, with B, etc., doesn't match terminology very well with the top one, which uses S', etc. Being consistent with node names, etc., would help!
You missed the sentences following 7:54 : "it is hard to identify...." . The circle at the op amp input is illustrative, it is NOT the summing error junction.
@@sambenyaakov
OK, I see. They're 2 different animals. I got it. Thanks very much.
I was wondering if you could remind me of something I can't find the rationale for....
I've read/heard it stated that the Fc crossover (0dB) must be such that Fc
A PWM system is a discrete sampled system so Nyquist sampling criteria holds. So this is the reason.
@@sambenyaakov doh!
Hi Professor.. can you please make a video on .
When to calculate phase margin with respect 0 degrees and 180 degrees...
It's very confusing atleast to me...
see ruclips.net/video/yJucyH3WOvU/видео.html
How much and lead time for the analyzer and transformer?
Please contact Omicron as I am not part of company.
you have """open loop bandwidth 8.33Khz"" not ""close loop bandwidth"" , is it ?? @ 25:21
close loop bandwidth shows how good this can compensate for load variation , if we have a load that changes around 50Khz , dose it still enable to make regulation at that frequency ??
how we can measure close loop bandwidth for switch mode converter ?
HI Paul. This is the (closed) loop-gain as in the title of slide 49. Then you have the open loop gains, and closed loop gains. For these you need to specify what is input and what is output. E.g. Gain between input to modulator and output. Or: closed loop gain between disturbance at the power input to output. So what do you mean by :" close loop bandwidth for switch mode converter "? Between what and what?
@@sambenyaakov Hi Sam , I see in slide 49 , it is written "loop gain " more exact and accurate is (closed) loop-gain. so this means close loop bandwidth between output to input is 8 Khz when the gain reach 0dB , so my question is if you have a load at output with 50Khz transient switching then this converter can't regulate it at that rate ? is it ? regulation bandwidth is not enough !
Try the RidleyBox instead.
Same price, but includes:
4-Channel Frequency Response Analyzer
4-channel Scope
Computer
Injection Isolator
Lifetime RidleyWorks License
Watch this video to see how design, implement in hardware AND measure a flyback all in hour.
ridleyengineering.com/videos-e/300-design-and-build-a-current-mode-controller-in-one-hour-webinar.html
Learn more about the RidleyBox at: www.ridleybox.com
Thank you for the few tips in this video.
I you ever have some time to answer my possibly silly questions, I have two :
- Is it a problem if the compensation is digital ? For an Infineon PMIC ic, I once asked the field engineer where to get the feedback signal in order to perform loop gain measurement (the datasheet internal schematic was not clear). He answered me that it is not possible to make such measurement because of digital loop. I've never understood this answer and don't see the reason why it would not be possible (an incorrect behavior when faced to a sinusoidal signal, maybe ?). I asked him why, and I'm still waiting for the answer since more than one year ^_^.
- Does it makes sense to perform the measurement with the "real load", i.e. on the end application board (different types of load, and continuously moving), rather than with loading resistors ? It seems weird to me to make such measurement (loop gain is load dependant). But a collegue of mine - an old experimented guy - always do this with rather good results (i.e. quickly get an optimized transient response).
Thanks again for being so active since many years with so much interesting subjects to share with us.
Thank inico.
-No problem to measure the loopgain on a digitally controlled converter. No need to get outside the controller. You can use the method I am showing in this video, injection in the feedback path. Always connected the excitation signal to a low impedance point like the output cap.
- Strictly speaking the measurement should be done with the real load. However a resistor is OK in most cases except for the case in which your inverter/converter is feeding another converter which might exhibit a negative dynamic resistance.
@@sambenyaakov Thank so much for your time at answering my questions. Waiting for your next topic (I still have so many to watch, though).
@@inico9366 👍